Question

2.56 gm of sulphur S8(s) is taken which is in equilibrium with its vapour according to reaction,
if vapours occupies 960ml at 1 atm and 273K then the degree of dissociation of Sg(s) will be (Given: R = 0.08]
0.35
0.45
0.55
0.11​

Answer 1

Answer:

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Answer 2

The degree of dissociation of $S_8(s)$ is 0.46

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $S_8$ = 2.56 g

Molar mass of $S_8$ = 256 g/mol

Putting values in above equation, we get:

$\text{Moles of }S_8=\frac{2.56g}{256g/mol}=0.01mol$

We are given:

Volume of vapor sulfur = 960 mL

At STP (at 273 K and 1 atm):

22.4 L of 22400 mL of volume is occupied by 1 mole of a gas

So, 960 mL of volume will be occupied by = $\frac{1}{22400}\times 960=0.043mol$ of sulfur gas

The chemical equation for the conversion of solid sulfur to vapor sulfur follows:

               $S_8(s)\rightleftharpoons 8S(g)$

Initial:        0.01

At eqllm:    0.01-x        8x

Evaluating the value of 'x'

$\Rightarrow 8x=0.043\\\\\Rightarrow x=\frac{0.043}{8}=0.0055$

So, amount of solid sulfur dissociated = (0.01 - x) = (0.01 - 0.0055) = 0.0045 moles

Calculating the degree of dissociation, $\alpha$ :

$\alpha=\frac{\text{Amount of solid sulfur dissociated}}{\text{Initial amount of solid sulfur}}$

$\alpha =\frac{0.0045}{0.01}=0.46$

Learn more about STP and number of moles:

https://brainly.com/question/13972001

https://brainly.com/question/919137

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